Which of the following is closest to the minimum voltage required to power the electrolysis of molten sodium chloride (NaCl) under standard conditions?

• A. 1.23 V
• B. 2.71 V
• C. 0.83 V
• D. 4.07 V

Answer: (D)

Verified step by step guidance

1. Identify the half-reactions occurring during the electrolysis of molten sodium chloride (NaCl). At the cathode, sodium ions (Na\^+) are reduced to sodium metal (Na), and at the anode, chloride ions (Cl\^-) are oxidized to chlorine gas (Cl\_2).
2. Write the half-reactions with their standard reduction potentials (E\^\circ): Cathode (reduction): \(\mathrm{Na\^+ + e^- \rightarrow Na}\) Anode (oxidation): \(\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}\) (reverse of the reduction half-reaction \(\mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-}\)).
3. Look up the standard reduction potentials for each half-reaction under standard conditions: - \(E\^\circ_{\mathrm{Na^+/Na}}\) is approximately -2.71 V, - \(E\^\circ_{\mathrm{Cl_2/Cl^-}}\) is approximately +1.36 V.
4. Calculate the overall cell potential (minimum voltage) required for electrolysis by subtracting the anode potential from the cathode potential: \(E_{\mathrm{cell}} = E\^\circ_{\mathrm{cathode}} - E\^\circ_{\mathrm{anode}}\). Since the anode reaction is oxidation, use the reduction potential of chlorine and reverse its sign for oxidation.
5. Sum the absolute values of the potentials to find the minimum voltage needed: \(E_{\mathrm{cell}} = |E\^\circ_{\mathrm{Na^+/Na}}| + |E\^\circ_{\mathrm{Cl_2/Cl^-}}|\). This value corresponds to the minimum voltage required to drive the electrolysis of molten NaCl under standard conditions.